Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 13}{x + 8} = \dfrac{-11x - 11}{x + 8}$
Solution: Multiply both sides by $x + 8$ $ \dfrac{x^2 + 13}{x + 8} (x + 8) = \dfrac{-11x - 11}{x + 8} (x + 8)$ $ x^2 + 13 = -11x - 11$ Subtract $-11x - 11$ from both sides: $ x^2 + 13 - (-11x - 11) = -11x - 11 - (-11x - 11)$ $ x^2 + 13 + 11x + 11 = 0$ $ x^2 + 24 + 11x = 0$ Factor the expression: $ (x + 8)(x + 3) = 0$ Therefore $x = -8$ or $x = -3$ At $x = -8$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -8$, it is an extraneous solution.